Задача 51897 ...

sardor51819

2020-05-29 09:25:02

∫ dx/((x+1)sqrt(x^2-3x+2))

sova

2020-05-29 10:06:29

★

Замена переменной:

[m]\frac{1}{x+1}=t[/m] ⇒ [m]x=\frac{1}{t}-1[/m]

[m]x^2-3x+2=(\frac{1}{t}-1)^2-3\cdot (\frac{1}{t}-1)+2=...=\frac{6t^2-5t+1}{t^2}[/m]

и

[m]dx=-\frac{dt}{t^2}[/m]
Тогда

[m] ∫ \frac{dx}{(x+1)\sqrt{x^2-3x+2}}[/m] = - ∫ [m]\frac{dt}{\sqrt{6t^2-5t+1}}[/m]


Выделяем полный квадрат:

[m]6t^2-5t+1=6(t^2-\frac{5}{6}t+\frac{1}{6})=6((t-\frac{5}{12})^2-\frac{1}{144})[/m]


[m]=- \frac{1}{\sqrt{6}}∫\frac{dt}{\sqrt{(t-\frac{5}{12})^2-\frac{1}{144}}}=[/m]

cм. формулу в приложении

[m]=- \frac{1}{\sqrt{6}}ln|t-\frac{5}{12}+\sqrt{t^2-\frac{5}{6}t+\frac{1}{6}}|+C[/m]

где [m]t=\frac{1}{x+1}[/m]